Variance of product of Brownian motions

Question

Let $\{B_{t}\}_{t\geq0}$ be Brownian motion. What is the variance of $B_{t}B_{s}$?

Answer 1

Assuming that $t\geq s$, write $B_t B_s=(B_t-B_s)B_s+B_s^2$. Taking the expectation, we find that $\mathbb{E}(B_t B_s)=s$. On the other hand $$(B_t B_s)^2=(B_t-B_s)^2B_s^2+2(B_t-B_s)B_s^3+B_s^4,$$ so taking expectation this time gives $$\mathbb{E}((B_t B_s)^2)=(t-s)s+0+3s^2.$$

Finally, taking the difference of these we get $$\mbox{Var}(B_t B_s)=\mathbb{E}((B_t B_s)^2)-\mathbb{E}(B_t B_s)^2=(t+s)s.$$